I want to solve this limit:
lim
I have proved that \lim\limits_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n} = 0 and \lim\limits_{n \rightarrow +\infty} \frac{1}{(1-\cos(1/n^2))}= \infty but I have indeterminate form. How can I solve that?
Answer
A long version:
\lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))}= \lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-\ln(n^3)}{n\cdot\frac{1}{n^4}\cdot\frac{(1-\cos^2(1/n^2))}{\frac{1}{n^4}}}\cdot(1+\cos(1/n^2))=\\ \lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-\ln(n^3)}{\frac{1}{n^3}\cdot \color{red}{\frac{\sin^2(1/n^2)}{\frac{1}{n^4}}}}\cdot(1+\cos(1/n^2))=\\ 2\cdot \lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-\ln(n^3)}{\frac{1}{n^3}}= 2\cdot \lim_{n \rightarrow +\infty} n^3\ln\left(\frac{1+n+n^3}{n^3}\right)=\\ 2\cdot \lim_{n \rightarrow +\infty} \ln\left(1+\frac{1+n}{n^3}\right)^{n^3}=\\ 2\cdot \lim_{n \rightarrow +\infty} \ln\left(1+\frac{1+n}{n^3}\right)^{\frac{n^3}{n+1}\cdot (n+1)} = \\ 2\cdot \lim_{n \rightarrow +\infty} (n+1)\cdot\ln\left(1+\frac{1+n}{n^3}\right)^{\frac{n^3}{n+1}} = 2\cdot \ln{e} \cdot \lim_{n \rightarrow +\infty} (n+1)\rightarrow +\infty
On the 2nd line
\lim_{n \rightarrow +\infty}\frac{\sin^2(1/n^2)}{\frac{1}{n^4}}= \lim_{n \rightarrow +\infty}\frac{\sin(1/n^2)}{\frac{1}{n^2}}\cdot \frac{\sin(1/n^2)}{\frac{1}{n^2}}=1 from
\lim\limits_{x\rightarrow0}\frac{\sin x}{x}=1
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